# Math Help - Boolean algebra help

1. ## Boolean algebra help

Hi,

I have a boolean function that im not sure how to solve.

$(x+z')'*(y+z')' + xy' +xz'$

im supposed to write this on disjunctive normal form.

I tried multiplying the $xy'$ and the $xz'$ term
by (z+z') and (y+y') but that makes the whole expression really long.

Any ideas?

Jones

2. ## A start to rewriting this

I assume (x+z')' means not ( x or (not z)). If so, then doesn't this simplify to
x'z? Then your expression simplifies to:

= x'zy'z + xy' + xz'
= x'y'z + xy' + xz'

3. No, thats incorrect.

The correct answer should be $x'y'z + xy'z' + xy'z + xyz'$

4. ## A little more manipulation

Note:
xy' = xy'z + xy'z'
because z + z'=1

5. So,

Simplifying the first two paranthesis
$\overline{x+\overline{z}} \ast \overline{y+\overline{z}}$

Should simplify down to just $\overline{x}z \ast \overline{y}z$

Applying Morgans law gives:

$\overline{\overline{x}z \ast \overline{y}z} \rightarrow x+\overline{z} + y+\overline{z}$

Using de morgans again gives: $\overline{x}z+\overline{y}z$

The whole expression now becomes $\overline{x}z+\overline{y}z+x\overline{y}+x\overli ne{z}$

Is this correct?

6. you need my advice it is a 3 literals function so it is easy to build the truth table and see where the function set to 1 then build the karnough map and you will get the most simplified form

7. the simplified form is z'x+zy'

truth table :
X Y Z F
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 0

K-map :
xy 00 01 11 10
z 0 0 0 (1 1)
1 1) 0 0 (1
=> z'x+zy'